topic 4 - traffic characterization

Traffic Characterization

Dr. Christos Drakos

University of Florida

Topic 4 – Traffic Characterization

1. IntroductionTraffic is the most important factor in pavement design; thickness is based on the number of load repetitions (traffic)

a. Fixed Traffic:• Thickness governed by single-wheel load (use the highest anticipated

load for design)• Used for heavy load / low volume pavements (i.e. airfields)

b. Fixed Vehicle Traffic:• Thickness governed by # of repetitions of a standard vehicle or axle-

load• Convert all traffic to 18-kip single axle loads

c. Variable Traffic and Vehicle:• Loads are divided into groups (load spectra) and the corresponding

stresses and strains are used for design• More appropriate for mechanistic design methods

1.1 Traffic Characterization Procedures


Fixed Vehicle Traffic• Design is based on the total number of passes of the standard

axle load (18-kip Equivalent Single Axle Load – ESAL) during the design period

• Covert all traffic to the standard axle load (ESAL)

How many ESAL does a 24-kip axle amount to?

( ) ( )3.291 0.854

f t 1N 0.0796 E− −

= ε

( ) 4.4779d cN 1.365 10

−−= × εNd, Nf = Load cycles to failure

= ESAL

Basic premise:Must determine how many 18-kip single axle loads would cause the same damage as one X-kip load


1.2 Equivalent Axle Load Factor (EALF) a.k.a. (LEF)• Defines the damage per pass to pavement by the axle in

question relative to the damage per pass of a standard 18-k axle

Load Equivalence Factor (LEF) Depends on:• Type of pavement• Thickness / structural capacity• Terminal conditions (definition of failure)

– 20% of lane area with fatigue cracking– ½ inch rutting

• Theoretical analysis Nf(18)/Nf(X)

• Based on experience (AASHO Road Test)– Table 6.4 (flexible pavements)– Table 6.7 (rigid pavements)


For the same structure apply 24- & 18-kip load

• KENLAYER:– (18-Kip) εt = 200µε → Nf(18) = 1,612,000– (24-Kip) εt = 267µε → Nf(24) = 623,000

• So, we can get an equivalent damage factor• Nf(18)/Nf(24) = 2.59• It would take 2.59 18-kip load single axles to cause the same

damage as one 24-kip axle

24-kip

AC

BASE

SUBGRADE

∞

18-kip

AC

BASE

SUBGRADE

∞

1.3 Theoretical Analysis to get LEF (Mechanistic)


LEF = No. of 18k Single Axle Load to cause specific damage

No. of Xk Single Axle Load to cause specific damage

1.3 Theoretical Analysis to get LEF (Mechanistic)

Due to the many factors that influence the LEF, it is almost impossible to select an appropriate a single value that applies to all situations. For a truly mechanistic design method, each load group should be analyzed separately.

Issues with theoretical analysis:• Does the LEF change if we modify structural configuration

(thickness, modulus, etc.)?• Which one is more critical – fatigue cracking or rutting

analysis?


[ ] 33.4x2G

G79.4

x2x

s218

18

x L10

10LLLL

WW

18

x

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡⎥⎦

⎤⎢⎣

⎡++

=β

β

Where:Wx = axle applications inverse of equivalency factors W18 = No of 18kip single axle loadsLx = axle load being evaluated (kips)L18 = 18 (standard axle load in kips)L2s = code for standard axle = 1 (single axle)L2x = code for axle load being evaluated

L2x = 1 for single axleL2x = 2 for tandem axleL2x = 3 for triple axle (added in the 1986 AASHTO Guide)

1.3 AASHTO Equivalent Factors (Empirical)

x

18

WWEALF =


function of the ratio of loss in serviceability at time, t, to the potential loss taken at a point where pt = 1.5⎥⎦

⎤⎢⎣⎡

−−

=5.12.4

p2.4LogG t

( )( ) ⎥

⎦

⎤⎢⎣

⎡+

++=β 23.3

x219.5

23.3x2x

L1SNLL081.04.0 function which determines the relationship

between serviceability and axle load applications

Where:pt = "terminal" serviceability index (point at which the

pavement is considered to be at the end of its useful life)

1.3 AASHTO Equivalent Factors (Empirical)

[ ] 33.4x2G

G79.4

x2x

s218

18

x L10

10LLLL

WW

18

x

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡⎥⎦

⎤⎢⎣

⎡++

=β

β

x

18

WWEALF =


1.4 AASHTO Equivalent Factors – Example 1Calculate the LEF for a 30,000 lb single-axle load. The structural number (SN) is equal to three (3) and the terminal serviceability is 2.5.

L18 18:= L2s 1:=

W18 = predicted number of 18-kip single axle load applicationsWx = predicted number of 30-kip single axle load applicationsLx = L 30 = 30L2x = 1 (single axle)

Lx 30:= pt 2.5:=

L2x 1:= SN 3:=

β18 0.40.081 L18 L2s+( )3.23

SN 1+( )5.19 L2s3.23⋅

⎡⎢⎢⎣

⎤⎥⎥⎦

+:= βx 0.40.081 Lx L2x+( )3.23

SN 1+( )5.19 L2x3.23⋅

⎡⎢⎢⎣

⎤⎥⎥⎦

+:=G log4.2 pt−

4.2 1.5−

⎛⎜⎝

⎞⎠

:=

EALFW18Wx

WxW18

L18 L2s+

Lx L2x+

⎛⎜⎝

⎞

⎠

4.7910

Gβ x

10

Gβ 18

⎛⎜⎜⎜⎜⎝

⎞

⎟⎟

⎠

L2x( )4.33

EALFL18 L2s+

Lx L2x+

⎛⎜⎝

⎞

⎠

4.7910

Gβ x

10

Gβ 18

⎛⎜⎜⎜⎜⎝

⎞

⎟⎟

⎠

L2x( )4.33

⎡⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎦

1−

:= EALF 7.935=


1.5 AASHTO Equivalent Factors – Example 2Calculate the LEF for a 40,000 lb tandem-axle load. The structural number (SN) is equal to f ive (5) and the terminal serviceability is 2.5.

L18 18:= L2s 1:=

W18 = predicted number of 18-kip single axle load applicationsWx = predicted number of 40-kip tandem axle load applicationsLx = L 40 = 40L2x = 2 (tandem axle)

Lx 40:= pt 2.5:=

L2x 2:= SN 5:=

β18 0.40.081 L18 L2s+( )3.23

SN 1+( )5.19 L2s3.23⋅

⎡⎢⎢⎣

⎤⎥⎥⎦

+:= βx 0.40.081 Lx L2x+( )3.23

SN 1+( )5.19 L2x3.23⋅

⎡⎢⎢⎣

⎤⎥⎥⎦

+:=G log4.2 pt−

4.2 1.5−

⎛⎜⎝

⎞⎠

:=

EALFW18Wx

WxW18

L18 L2s+

Lx L2x+

⎛⎜⎝

⎞

⎠

4.7910

Gβ x

10

Gβ 18

⎛⎜⎜⎜⎜⎝

⎞

⎟⎟

⎠

L2x( )4.33

EALFL18 L2s+

Lx L2x+

⎛⎜⎝

⎞

⎠

4.7910

Gβ x

10

Gβ18

⎛⎜⎜⎜⎜⎝

⎞

⎟⎟

⎠

L2x( )4.33

⎡⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎦

1−

:= EALF 2.081=


• Where:– ADT0 = Initial Average Daily Traffic– T = Percent Trucks (decimal)– TF = Truck Factor (decimal)– G = Growth Factor– Y = Design Period– L = Lane Distribution Factor (decimal)– D = Directional Distribution Factor (decimal)

ESAL = Equivalent Single Axle LoadESALs = Cumulative ESALs for all vehicles over the entire design period (we can also calculate ESAL for specific vehicle type)

2. Computation of Design ESALs

ESALs = (ADT0)(T)(TF)(G)(Y)(D)(L)*365


2. Computation of Design ESALsDesign ESALs:

∑=

=m

1i(i)ESALESALs m = vehicle types Must calculate ESALs for

each vehicle type

2.1 Average Daily Traffic (ADT) Unless otherwise stated, ADT is in all lanes & both directionsAlso, ADT includes:

• Cars• Single-unit trucks & buses• Multiple-unit trucks

ESAL(i) = (ADT0)(T)(Ti)(TFi)(G)(Y)(D)(L)*365

New terms:Ti = Distribution of specific type of truck within all trucks (decimal)TFi = Truck factor for the specific truck type (decimal)


2.2 Average Daily Truck Traffic (ADTT) or (T)• Minimum traffic information required for pavement design;

everything else can be found in tables• Very important in pavement design

– Effort to collect actual data– Table 6.9; guide to distribution of truck types among total amount of

trucks

2.2.1 Example:• 4000 ADT ; 20% Trucks; Rural System/Principal• Find the # of 2-axle, 4-tire trucks

# 2-axle, 4-tire trucks = (4000) (0.2) (365)

ADT T



2.2 Average Daily Truck Traffic (ADTT) or (T)

• Minimum traffic information required for pavement design; everything else can be found in Tables

• Very important in pavement design– Effort to collect actual data– Table 6.9; guide to distribution of truck types among total

amount of trucks

# 2-axle, 4-tire trucks = (4000) (0.2) (365)

ADT T

(0.6)

Table 6.9

=175,000/year

2.2.1 Example:• 4000 ADT ; 20% Trucks; Rural System/Principal• Find the # of 2-axle, 4-tire trucks


2.3 Truck Factor (TF)

TF = ESALs# of Trucks

Sum of ESALs divided by the number of trucks weighed (count of trucks, not axles)

What is the importance of TF?

• Single TF can be applied to all trucks (weighed average); or separate for each truck type if the growth rates are different

• Table 6.10 (Truck Factors)– If we use all trucks, we do not have to calculate ESALs

for each truck type

For the same ESALs; if TF increases Less # of trucks

If less # of trucks produce the same ESALs More severe loads



2.4 Directional Distribution (D)• Usually assume D = 0.5

• Function of ADT & # of lanes (Table 6.14)

Where could that be different?

2.5 Lane Distribution (L)

Outer traffic

Center traffic

Inner traffic

• We design for Outer Lane, but everything is built the same

• Inner Lane usually under-loaded


2.6 Growth Rate Factor (G)

• Assuming a yearly rate of growth (r)

( )[ ]Yr1121

G ++×=

• Asphalt Institute (Table 6.13); (G)(Y) combined


4-lane Rural/Principal4000 ADT20% Trucks20-year design; r=4%

Determine the ESALs for 2-axle, 6-tire trucks

ESAL(i) = (ADT0)(Ti)(TFi)(G)(Y)(D)(L)*365

• Distribution of trucks (Table 6.9)– Ti = 10% (for 2-axle, 6-tire) * 20% (trucks) = 0.02

• Truck factor (Table 6.10)– TFi = 0.25

• Growth (Table 6.13)– GY = 29.78

• Lane Distribution (Table 6.14)– L = 0.94 ESAL(i)=4000*0.02*0.25*29.87*0.5*0.94*365

= 102,175 ESALs/20 years

2.7 Example 1


4-lane Rural/Principal4000 ADT20% Trucks20-year design; r=4%

Determine the total ESALs

ESALs = (ADT0)(T)(TF)(G)(Y)(D)(L)*365

• Trucks – T = 20% (trucks) = 0. 2

• Truck factor (Table 6.10)– TF = 0.38

ESALs = 4000*0.2*0.38*29.87*0.5*0.94*365= 1.55*106 ESALs/20 years

2.8 Example 2

topic 4 - traffic characterization

Documents